Math Twelve

Determinants NCERT Solutions

NCERT Exercise 4.5

Find adjoint of each of the matrices in Exercise 1 and 2.

Question:1. `|(1, 2), (3, 4)|`

Solution:

Let `|(1, 2), (3, 4)|`

`:. A_11=4,A_12=` `-2,A_22+1`

`:.adj A=|(4, -3), (-2, 1)|=` `|(4, -2), (-3, 1)|`

Question:2. `|(1, -1, 2), (2, 3, 5), (-2, 0, 1)|`

Solution:

`|(1, -1, 2), (2, 3, 5), (-2, 0, 1)|`

`A_11=(3-0)=3, A_12`

`=-(2+10)=-12, A_13`

`=(0+6 )=6,A_21`

`=-(-1-0)=1`

`A_22-(1+4)`

`=5,A_23=-(0-2)`

`=2,A_31=(-5-6)`

`=-11,A_32=`

`=-(-5-4)=-1`

`A_33=(3+2)=5`

`:.adj A = [(3, -12, 6), (1, 5, 2), (-11, -1, 5)]`

`=[(3, 1, -11), (-12, 5, -1), (6, 2, 5)]`

Verify A (adj A)A=|A| I in Exercises 3 and 4

Question:3. `[(2, 3), (-4, -6)]`

Solution:

Here A=`[(2, 3), (-4, -6)]`

Now |A|=`|(2, 3), (-4, -6)| `

`=-12-(-12)`

`=-12+12=0`

`A_11=-6,A_12`

`=-(-4)=4, A_21`

`=-3,A_22=2`

`:. adj A= [(-6, 4), (-3, 2)]`

`=[(-6, -3),(4, 2)]`

`A(adj A)=[(2, 3),(-4, -6)]` `[(-6, -3), (4, 2)]`

` =[(-12+12, -6+6), (24-24, 12, 12)]`

`=[(0, 0), (0, 0)]= 0`

`(adj A)A = [(-6, -3), (4, 2)]` `[(2, 3), (-4, -6)]

`=[(-12+12, -18+18),(8-8, 12-12)]`

`=[(0, 0),(0, 0)]=0`

`|A| I =0[(1, 0), (0, 1)]=0`

Thus `A(adj A)=(adj A)A =|A|I`

Question:4 .

`[(1, -1, 2),(2, 3, 5),(-2, 0, 1), (1, 0, 3)]`

Solution:

Let `[(1, -1, 2),(2, 3, 5),(-2, 0, 1), (1, 0, 3)]`

`|A|=[(1, -1, 2),(2, 3, 5),(-2, 0, 1), (1, 0, 3)]`

`=1(0-0)-(-1)` `(9+2)+ 2(0-0)

`=0 +11+ 0=11`

`:. A_11 + (0-0)` `0,A_12=-(9+2)`

`=-11,A_13=(0-0)=0`

`A_21=-(-3-0)`

`3, A_22=(3-2)`

`=1, A_23`

`=-(0+1)=-1`

`A_31(2-0)`

`=2,A_32`

`=-(-2-6)`

`=8,A_33`

`=(0+3)=3`

`adj A=[(0, -11, 0), (3, 1, -1), (2, 8, 3)]`

`=[(0, 3, 2), (-11, 1, 8), (0, -1, 3)]`

`A(adj A)` `= [(1, -1, 2),(3, 0, -2), (1, 0, 3)][(0, 3, 2),(-11, 1, 8),(0, -1, 3)]`

`[(0+11+0, 3-1-2, 2-8+6),(0+0+0, 9+0+2, 6+0-6), (0+0+0, 3+0-3, 2+0+9)]`

`=[(11, 0, 0),(0, 11, 0),(0, 0, 11)]`

`=11[(1, 0, 0), (0, 1, 0),(0, 0, 1)]` `=|A|I_3`

`(adj A)` `=[(0, 3, 2), (-11, 1, 8), (0, -1, 3)]` `[(1, -1, 2), (3, 0, -2), (1, 0, 3)]`

`=[(0+9+2, 0+0s+0, 0+6+6), (-11+3+8, 11+0+0, -22-2+24), (0-3+3, 0+0+0, 0+2+9)]`

`=[(11, 0, 0), (0, 11, 0), (0, 0, 11)]`

`=11[(1, 0, 0), (0, 1, 0), (0, 0, 1)]`

`=|A|I_3`

Thus A` (adj A)A=|A|I_3`

Find the inverse of each of the matrices(if it exists)given in Exercises5 to 11

Question:5. `[(2, -2), (4, 3)]`

Solution:

Let`[(2, -2), (4, 3)]`

|A|=`[(2, -2), (4, 3)]`

`=6-(-8)` `=6+8=14`

`A_11=3,A_12` `=-4,A+21`

`=-(-2)=2,A_22=2`

`adj A= [(3, -4), (2, 2)]`

`=[(3, 2), (-4, 2)]`

`A^-1 = 1/|A| adj A`

`1/(14)[(3, 2), (-4, 2)]`

Question:6 . `[(-1, 5), (-3, 2)]`

Solution:

Let`[(-1, 5), (-3, 2)]`

`|A|[(-1, 5), (-3, 2)]`

`=-2-(-15)`

`=-2+15=13`

`A_11=2,A_12=`

`=-(-3)=3,A_21`

`=-5,A_22=-1`

`adj A=[(2, 3), (-5, -1)]`

`=[(2, -5), (3, -1)]`

`:. A^-1 = 1/|A| adj A`

`=1/13 [(2, -5), (3, -1)]`

Question:7 . `[(1, 2, 3), (0, 2, 4), (0, 0, 5)]`

Solution:

Let`[(1, 2, 3), (0, 2, 4), (0, 0, 5)]`

`|A|[(1, 2, 3), (0, 2, 4), (0, 0, 5)]`

`=1(10-0)-2(0-0)` `+3(0-0)`

`10-0+0=10`

`A_11=(10-0)=10,A_12`

`=-(0-0)=0,A_13` `=(0-0)=0`

`A_21=(10-0)=-10,A_22`

`=(5-0)=5,A_23` `=-(0-0)=-(0-0)=0`

`A_31=(8-6)=2,A_32`

`=-(4-0)=-4,A_33` `=(2-0)=2`

`adj A=[(10, 0, 0),(-10, 5, 0), (2, -4, 2)]`

`=[(10, -10, 2),(0, 5, -4), (0, 0, 2)]`

`A^-1 =1/|A| adj A`

`1/10 [(10, -10, 2),(0, 5, -4), (0, 0, 2)]`

Question:8 . `[(1, 0, 0), (3, 3, 0), (5, 2, -1)]`

Solution:

Let A =`[(1, 0, 0), (3, 3, 0), (5, 2, -1)]`

`|A|=[(1, 0, 0), (3, 3, 0), (5, 2, -1)]`

`1(-3-0)-0(-3-0)` `+0(6-15)`

`=-3-0+0=-3`

`A_11=(-3-0)`

`=-3,A_12`

`=-(-3-0)`

`= 3, A_13`

`=(6-15)=-9`

`A_21=-(0-0)0,A_22`

`=(1-0)=-1,A_23`

`=-(2-0)=-2`

`A_31=(0-0)=0,A_32`

`=-(0-0)=0, A_33`

`=(3-0)=3`

`adj A[(-3, 3, -9), (0, -1, -2), (0, 0, 3)]`

`= [(-3, 0, 0), (3, -1, 0), (-9, -2, 3)]`

`A^-1=1/|A|adj A`

`=1/-3[(-3, 0, 0), (-3, 1, 0), (9, 2, -3)]`

`1/3= [(3, 0, 0), (-3, 1, 0), (9, 2, -3)]`

12-math-home


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